Determinant left| {begin{array}{*{20}{c}}{1, + ,a, + ,x}&{a, + ,y}&{a, + ,z}{b, + ,x}&{1

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3 and 4 .Determinants and Matrices

normal

The determinant $\left| {\begin{array}{*{20}{c}}{1\, + \,a\, + \,x}&{a\, + \,y}&{a\, + \,z}\\{b\, + \,x}&{1\, + \,b\, + \,y}&{b\, + \,z}\\{c\, + \,x}&{c\, + \,y}&{1\, + \,c\, + \,z}\end{array}} \right|$ $=$

$(1 + a + b + c) (1 + x + y + z) - 3 (ax + by + cz)$

$a (x + y) + b (y + z) + c (z + x) - (xy + yz + zx)$

$x (a + b) + y (b + c) + z (c + a) - (ab + bc + ca)$

none of these

Solution

$1 + a + b + c = k$ and use $R_1 + R_2 + R_3$ we get

$D =$ $\left| {\,\,\begin{array}{*{20}{c}}{k + 3x}&{k + 3y}&{k + 3z}\\{b + x}&{1 + b + y}&{b + z}\\{c + x}&{c + y}&{1 + c + z}\end{array}\,} \right|$

$= k$ $\left|{\,\begin{array}{*{20}{c}}1&1&1\\{b + x}&{1 + b + y}&{b + z}\\{c + x}&{c + y}&{1 + c +z}\end{array}\,} \right|$ +$3\,\,\,\left| {\,\begin{array}{*{20}{c}}x&y&z\\{b + x}&{1 + b + y}&{b + z}\\{c + x}&{c + y}&{1 + c + z}\end{array}\,} \right|$

now proceed

Standard 12

Mathematics

$\left| {\,\begin{array}{*{20}{c}}{b + c}& a& a\\b& {c + a}& b\\c& c& {a + b}\end{array}\,} \right| = $

medium

Using properties of determinants, prove that:

$\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$

hard

$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}\,} \right| = $

medium

If $D =$ $\left| {\,\begin{array}{*{20}{c}}{\frac{1}{z}}&{\frac{1}{z}}&{ – \frac{{(x + y)}}{{{z^2}}}}\\{ – \frac{{(y + z)}}{{{x^2}}}}&{\frac{1}{x}}&{\frac{1}{x}}\\{ – \frac{{y(y + z)}}{{{x^2}z}}}&{\frac{{x + 2y + z}}{{xz}}}&{ – \frac{{y(x +y)}}{{x{z^2}}}}\end{array}\,} \right|$ then, the incorrect statement is

normal

For a non – zero, real $a, b$ and $c$ $\left| {\begin{array}{*{20}{c}}{\frac{{{a^2} + {b^2}}}{c}}&c&c\\a&{\frac{{{b^2} + {c^2}}}{a}}&a\\b&b&{\frac{{{c^2} + {a^2}}}{b}} \end{array}} \right|$ $= \alpha \, abc$, then the values of $\alpha$ is

normal

The determinant $\left| {\begin{array}{*{20}{c}}{1\, + \,a\, + \,x}&{a\, + \,y}&{a\, + \,z}\\{b\, + \,x}&{1\, + \,b\, + \,y}&{b\, + \,z}\\{c\, + \,x}&{c\, + \,y}&{1\, + \,c\, + \,z}\end{array}} \right|$ $=$

Solution

Similar Questions

$\left| {\,\begin{array}{*{20}{c}}{b + c}& a& a\\b& {c + a}& b\\c& c& {a + b}\end{array}\,} \right| = $

Using properties of determinants, prove that: $\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$

$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}\,} \right| = $

For a non – zero, real $a, b$ and $c$ $\left| {\begin{array}{*{20}{c}}{\frac{{{a^2} + {b^2}}}{c}}&c&c\\a&{\frac{{{b^2} + {c^2}}}{a}}&a\\b&b&{\frac{{{c^2} + {a^2}}}{b}} \end{array}} \right|$ $= \alpha \, abc$, then the values of $\alpha$ is

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Using properties of determinants, prove that:

$\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$